Question

(i) If $\tan \left(A-B\right)=\frac{1}{\sqrt{3}}\mathrm{and}\tan \left(A+B\right)=\sqrt{3},0°<A+B\le 90°,A>B$ find A and B.

(ii) If $\tan \left(A+B\right)=1\mathrm{and}\tan \left(A-B\right)=\frac{1}{\sqrt{3}},0°<A+B<90°,A>B$, then find the values of A and B.

Answer 1

(i)

Given:

…… (1)

…… (2)

We know that,

…… (3)

…… (4)

Now by comparing equation (1) and (3)

We get,

…… (5)

Now by comparing equation (2) and (4)

We get,

…… (6)

Now to get the values of A and B, let us solve equation (5) and (6) simultaneously

Therefore by adding equation (5) and (6)

We get,

Therefore,

Hence

Now by subtracting equation (5) from equation (6)

We get,

Therefore,

Hence

Therefore the values of A and B are as follows

and

(ii)
$$\begin{gathered} \tan \left(A+B\right)=1 \\ \Rightarrow \tan \left(A+B\right)=\tan 45° \\ \Rightarrow A+B=45°.....\left(1\right) \end{gathered}$$

Also,
$$\begin{gathered} \tan \left(A-B\right)=\frac{1}{\sqrt{3}} \\ \Rightarrow \tan \left(A-B\right)=\tan 30° \\ \Rightarrow A-B=30°.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} A+B+A-B=45°+30° \\ \Rightarrow 2A=75° \\ \Rightarrow A=37.5° \end{gathered}$$

Putting $A=37.5°$ in (1), we get

$$\begin{gathered} 37.5°+B=45° \\ \Rightarrow B=45°-37.5° \\ \Rightarrow B=7.5° \end{gathered}$$

Thus, the values of A and B are 37.5º and 7.5º, respectively.